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JEE Mains Maths · Limits and Continuity

Limits at Infinity and Limits of Sums

Limits as x or n grows without bound: comparing the fastest-growing terms, summing a finite series before taking the limit, and turning a sum into an integral.

All Limits and Continuity notesJEE Mains Maths strategy

Why this matters

Seventeen PYQs, sixteen of them multiple choice. Six compare the highest powers or rationalise a difference of surds; five sum a series in closed form first; six recognise a Riemann sum and evaluate it as an integral. Three ideas cover the page.

Concept 1 of 3: The dominant terms

As x→∞x\to\infty, a polynomial behaves like its highest power, so a ratio of polynomials behaves like the ratio of leading terms. A difference like x2+x−x\sqrt{x^2+x}-x is ∞−∞\infty-\infty; rationalise it first, then compare leading terms.

Definition

  • Equal degrees: the ratio of leading coefficients.
  • Higher degree on top: ±∞\pm\infty; higher below: 0.
  • P−Q\sqrt{P}-\sqrt Q: multiply by the conjugate.
  • exe^x beats any power of xx, which beats ln⁡x\ln x.

Leading terms

lim⁡x→∞anxn+…bnxn+…=anbn\lim_{x\to\infty}\frac{a_nx^n+\dots}{b_nx^n+\dots}=\frac{a_n}{b_n}

Worked example

Find lim⁡x→∞3x2+52x2−x\lim_{x\to\infty}\frac{3x^2+5}{2x^2-x}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q156Moderate

Example 1 · Limits and Continuity · Limits at Infinity and Limits of Sums

lim⁡x→∞(3x+1+3x−1)6+(3x+1−3x−1)6(x+x2−1)6+(x−x2−1)6x3\lim_{x \rightarrow \infty} \frac{(\sqrt{3x + 1}+\sqrt{3x - 1})^{6}+ (\sqrt{3x + 1}-\sqrt{3x - 1})^{6}}{\left( x +\sqrt{x^{2}- 1} \right)^{6}+\left( x -\sqrt{x^{2}- 1} \right)^{6}}x^{3}

Infinity minus infinity is not 0

x2+x−x\sqrt{x^2+x}-x tends to 12\frac12, not 0. Rationalise before comparing.
Drill 5 more on the dominant terms

Concept 2 of 3: Sum first, then take the limit

When the number of terms grows with nn, the limit of the sum is not the sum of the limits. Find the sum in closed form — arithmetic or geometric series, ∑k\sum k, ∑k2\sum k^2, telescoping — and then let n→∞n\to\infty.

Definition

  • ∑k=1nk=n(n+1)2\sum_{k=1}^nk=\frac{n(n+1)}2, ∑k2=n(n+1)(2n+1)6\sum k^2=\frac{n(n+1)(2n+1)}6, ∑k3=(n(n+1)2)2\sum k^3=\left(\frac{n(n+1)}2\right)^2.
  • Telescoping: 1k(k+1)=1k−1k+1\frac1{k(k+1)}=\frac1k-\frac1{k+1}.
  • Geometric: ∑k≥0rk=11−r\sum_{k\ge0}r^k=\frac1{1-r} for ∣r∣<1|r|<1.

Power sums

∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6

Worked example

Find lim⁡n→∞1+2+⋯+nn2\lim_{n\to\infty}\frac{1+2+\dots+n}{n^2}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q65Moderate

Example 2 · Limits and Continuity · Limits at Infinity and Limits of Sums

The value of lim⁡n→∞(∑K=1nk3+6k2+11k+5(k+3)!)\lim_{n \rightarrow \infty} \left( \sum_{K = 1}^{n} \frac{k^{3}+ 6k^{2}+ 11k + 5}{(k + 3)!} \right) is:

Many small terms can add to something

Each term kn2\frac k{n^2} tends to 0, but there are nn of them and their sum tends to 12\frac12. Sum first.
Drill 4 more on sum first, then take the limit

Concept 3 of 3: Sums as integrals

A sum 1n∑k=1nf(kn)\frac1n\sum_{k=1}^nf\left(\frac kn\right) adds the areas of nn thin rectangles under y=f(x)y=f(x) on [0,1][0,1], so its limit is ∫01f(x) dx\int_0^1f(x)\,dx. Write the general term as 1n\frac1n times a function of kn\frac kn, then integrate.

Definition

  • lim⁡n→∞1n∑k=1nf(kn)=∫01f(x) dx\lim_{n\to\infty}\frac1n\sum_{k=1}^nf\left(\frac kn\right)=\int_0^1f(x)\,dx.
  • Replace kn\frac kn by xx and 1n\frac1n by dxdx.
  • If kk runs to 2n2n, the upper limit becomes 2.

Riemann sum

lim⁡n→∞1n∑k=1nf ⁣(kn)=∫01f(x) dx\lim_{n\to\infty}\frac1n\sum_{k=1}^{n}f\!\left(\frac kn\right)=\int_0^1f(x)\,dx

Worked example

Find lim⁡n→∞∑k=1n1n+k\lim_{n\to\infty}\sum_{k=1}^n\frac1{n+k}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q73Moderate

Example 3 · Limits and Continuity · Limits at Infinity and Limits of Sums

lim⁡n→∞[11+n+12+n+13+n+⋯+12n]\lim_{n \rightarrow \infty} \left\lbrack \frac{1}{1 + n}+\frac{1}{2 + n}+\frac{1}{3 + n}+ \cdots +\frac{1}{2n} \right\rbrack is equal to

Find the right limits of integration

The limits come from where kn\frac kn starts and ends. If kk runs from nn to 3n3n, the integral is over [1,3][1,3].
Drill 5 more on sums as integrals

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • The dominant terms

    Leading terms

    lim⁡x→∞anxn+…bnxn+…=anbn\lim_{x\to\infty}\frac{a_nx^n+\dots}{b_nx^n+\dots}=\frac{a_n}{b_n}
  • Sum first, then take the limit

    Power sums

    ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6
  • Sums as integrals

    Riemann sum

    lim⁡n→∞1n∑k=1nf ⁣(kn)=∫01f(x) dx\lim_{n\to\infty}\frac1n\sum_{k=1}^{n}f\!\left(\frac kn\right)=\int_0^1f(x)\,dx

Watch out for (3)

Test yourself on Limits and Continuity

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Limits and Continuity chapter testDrill all 17 questions